Problem: Solve for $x$ and $y$ using elimination. ${3x-5y = 14}$ ${3x-6y = 12}$
Solution: We can eliminate $x$ by adding the equations together when the $x$ coefficients have opposite signs. Multiply the bottom equation by $-1$ ${3x-5y = 14}$ $-3x+6y = -12$ Add the top and bottom equations together. ${y = 2}$ Now that you know ${y = 2}$ , plug it back into $\thinspace {3x-5y = 14}\thinspace$ to find $x$ ${3x - 5}{(2)}{= 14}$ $3x-10 = 14$ $3x-10{+10} = 14{+10}$ $3x = 24$ $\dfrac{3x}{{3}} = \dfrac{24}{{3}}$ ${x = 8}$ You can also plug ${y = 2}$ into $\thinspace {3x-6y = 12}\thinspace$ and get the same answer for $x$ : ${3x - 6}{(2)}{= 12}$ ${x = 8}$